Question

HEPES is a commonly used biochemical buffer with a pKa 7.5 at 25 °C. Please answer the questions below on the use of HEPES as a buffer. Use the Henderson-Hasselbalch equation.

a) What is the pH of a solution prepared by combining 100.0 mL of 0.50 M protonated HEPES with 50.0 mL of 0.20 M HEPES base and diluting to a final volume of 1.00 L?

b) You add 20 mL of 1.0 M NaOH to the solution prepared in Part a. What is the final pH and concentration of all relevant HEPES species in the resultant solution?

Answer 1

Sol.

(a) Acid = protonated HEPES

Salt = HEPES base

As moles of acid = Conc. of Acid × Volume of Acid / 1000

= 0.50 × 100 / 1000 = 0.05 mol

and , moles of salt = Conc. of salt × Volume of salt / 1000

= 0.20 × 50 / 1000 = 0.01 mol

As Volume of solution = 1 L

So , Conc. of acid in solution = [Acid]  

= 0.05 / 1 = 0.05 M

and , Conc. of salt in solution = [Salt]

= 0.01 / 1 = 0.01 M  

As pKa = 7.5  

So , Using Henderson - Hasselbalch equation ,

pH = pKa + log ( [Salt] / [Acid] )

= 7.5 + log ( 0.01 / 0.05 )

= 6.80

(b) As Moles of NaOH added = Conc. of NaOH × Volume of NaOH / 1000 = 1.0 × 20 / 1000 = 0.02 mol

So , moles of salt increases and moles of acid decreases

So , Final moles of salt = 0.01 + 0.02 = 0.03 mol

Final moles of acid = 0.05 - 0.02 = 0.03 mol

As Final  Volume of solution

= 1000 + 20 = 1020 mL  

= 1.02 L

Therefore , Final  Conc. of salt = [Salt] =   0.03 / 1.02 = 0.0294 M

and Final Conc. of acid = [Acid] = 0.03 / 1.02 = 0.0294 M

Also ,

Final pH = pKa + log ( [Salt] / [Acid] )

= 7.5 + log ( 0.0294 / 0.0294 )

= 7.5